Problem: Find the equation of the directrix of the parabola $y = 8x^2 + 2.$
Recall that a parabola is defined as the set of all points that are equidistant to the focus $F$ and the directrix.  To make the algebra a bit easier, we can find the directrix of the parabola $y = 8x^2,$ and then shift it upward 2 units to find the directrix of the parabola $y = 8x^2 + 2.$

Since the parabola $y = 8x^2$ is symmetric about the $y$-axis, the focus is at a point of the form $(0,f).$  Let $y = d$ be the equation of the directrix.

[asy]
unitsize(1.5 cm);

pair F, P, Q;

F = (0,1/4);
P = (1,1);
Q = (1,-1/4);

real parab (real x) {
  return(x^2);
}

draw(graph(parab,-1.5,1.5),red);
draw((-1.5,-1/4)--(1.5,-1/4),dashed);
draw(P--F);
draw(P--Q);

dot("$F$", F, NW);
dot("$P$", P, E);
dot("$Q$", Q, S);
[/asy]

Let $(x,8x^2)$ be a point on the parabola $y = 8x^2.$  Then
\[PF^2 = x^2 + (8x^2 - f)^2\]and $PQ^2 = (8x^2 - d)^2.$  Thus,
\[x^2 + (8x^2 - f)^2 = (8x^2 - d)^2.\]Expanding, we get
\[x^2 + 64x^4 - 16fx^2 + f^2 = 64x^4 - 16dx^2 + d^2.\]Matching coefficients, we get
\begin{align*}
1 - 16f &= -16d, \\
f^2 &= d^2.
\end{align*}From the first equation, $f - d = \frac{1}{16}.$  Since $f^2 = d^2,$ $f = d$ or $f = -d.$  We cannot have $f = d,$ so $f = -d.$  Then $-2d = \frac{1}{16},$ so $d = -\frac{1}{32}.$

Thus, the equation of the directrix of $y = 8x^2$ is $y = -\frac{1}{32},$ so the equation of the directrix of $y = 8x^2 + 2$ is $\boxed{y = \frac{63}{32}}.$